Problem: Solve for $r$, $ -\dfrac{r - 1}{6r} = \dfrac{8}{6r} + \dfrac{1}{6r} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $6r$ $6r$ and $6r$ The common denominator is $6r$ The denominator of the first term is already $6r$ , so we don't need to change it. The denominator of the second term is already $6r$ , so we don't need to change it. The denominator of the third term is already $6r$ , so we don't need to change it. This give us: $ -\dfrac{r - 1}{6r} = \dfrac{8}{6r} + \dfrac{1}{6r} $ If we multiply both sides of the equation by $6r$ , we get: $ -r + 1 = 8 + 1$ $ -r + 1 = 9$ $ -r = 8 $ $ r = -8$